Locally Sierpinski Space

Question

A topological space X is called a locally Sierpinski space if for all $x \in X$ there is an open neighbourhood of x homeomorphic to the Sierpinski space. It is usually denoted LS-space. $\forall x \in X$ there is exists $ S_{x} $ = $\lbrace x, y\rbrace$, $ x \neq y $, called the Sierpinski set, such that the relative topology on $ S_{x} $ is copy of the Sierpinski topology. How do i show that the Locally Sierpinski SPace is Alexandorff space ( $ \tau_{0} $) Help me to prove it using the DEFINITION: Let (X, $\tau$) be a topological space. X is said to be Kolmogorov (T$_{0}$) if for all x, y $\in$ X where x $\neq$ y, there exists some O $\in$ $\tau$ such that $\mid$ O $\bigcap$ $\lbrace{x,y}\rbrace$ $\mid$ = 1.

Answer 1

Recall that Sierpiński space is a two point space with one non-trivial open singleton set, the other singleton is closed but not open.

If $X$ is locally Sierpiński, let $x \neq y$ be two points of $X$. Let $L_x=\{x,a\}$ be the promised copy of Sierpiński space that is open in $X$ and contains $x$. If $\{x\}$ happens to be the open singleton in $L_x$, then as $L_x$ is open in $X$, so is $O=\{x\}$ and this $O$ is as required, or otherwise $a$ is the open singleton and now we have two cases: $a=y$ in which case $O=\{y\}$ is as required, or $a \neq y$ (so in fact $y \notin L_x$) and $O=L_x$ itself will do. $O$ contains one point of $\{x,y\}$ in all subcases.

So $X$ is $T_0$.